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Marat540 [252]
3 years ago
5

f a stadium pays $11,000 for labor and $7,000 for parking, what would the stadium's parking revenue have to be if the stadium is

hoping parking labor is 6% of parking revenue? a) $116,666 b) $183,333 c) $189,400 d) $190,000
Physics
2 answers:
Svetlanka [38]3 years ago
8 0

Answer: B

183,333

Explanation:

because i said so

QveST [7]3 years ago
7 0

Answer:

Explanation:

a

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A force of 5 N is applied to the end of a lever that has a length
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The answer is 10Nm

Explanation: I ended up just messing around with the numbers, I multiplied 5 and 2 got 10 as my answer and it was right.

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A space shuttle sits on the launch pad for 2.0 minutes, and then goes from rest to 4600 m/s in 8.0 minutes. Treat its motion as
SpyIntel [72]

Answer:

a.) a = 0 ms⁻²

b.) a = 9.58 ms⁻²

c.) a = 7.67 ms⁻²

Explanation:

a.)

    Acceleration (a) is defined as the time rate of change of velocity

                       a = \frac{v_{2} - v_{1} } {t}  

Given data

 Final velocity = v₂ = 0 m/s

 Initial velocity = v ₁ = 0 m/s

  As the space shuttle remain at rest for the first 2 minutes i.e there is no change in velocity so,

                 a = 0 ms⁻²

b.)

     Given data

As the space shuttle start from rest, So initial velocity is zero

    Initial velocity = v₁ = 0 ms⁻¹

    Final velocity  = v₂ = 4600 ms⁻¹

     Time = t = 8 min = 480 s

By the definition of Acceleration (a)

             a = \frac{v_{2} - v_{1} } {t}  

             a = \frac{4600 - 0 } {480}

                     a = 9.58 ms⁻²

c.)

    Given data

As the space shuttle is at rest for first 2 min then start moving, So initial velocity is zero

    Initial velocity = v₁ = 0 ms⁻¹

    Final velocity  = v₂ = 4600 ms⁻¹

     Time = t = 10 min = 600 s

By the definition of Acceleration (a)

             a = \frac{v_{2} - v_{1} } {t}  

             a = \frac{4600 - 0 } {600}

                     a = 7.67 ms⁻²

8 0
3 years ago
A constant force of 12N is applied for 3.0s to a body initially at rest. The final velocity of the body is 6.0ms–1. What is the
sp2606 [1]
From the question,
u = 0m {s}^{ - 1}
v = 6m {s}^{ - 1}

t = 3s
F=12N



Using Impulse, the product of the constant force, F and time t equals the product of the mass of the body and change in velocity.

Ft =m(v-u)


12(3.0)=m(6.0- \: 0)
This implies that

36.0 = 6m
m =  \frac{36.0}{6.0}
\therefore \: m = 6.0kg


You can also use the equation of linear motion,
v = u + at
6 = 0 + a(3)
6 = 3a
a =  \frac{6}{3}

a = 2 {ms}^{ - 2}
But
F=ma
12 = m(2)
12 = 2m
\frac{12}{2}  = m
\therefore \: m = 6kg
4 0
3 years ago
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C.) Monsoons.
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3 years ago
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